Determinants and $2\times 2$ minors I just proved that 
$$0=\left|\begin{array}{cccc}
 a_0 & a_1 & a_2 & a_3 \\
 b_0 & b_1 & b_2 & b_3 \\
 a_0 & a_1 & a_2 & a_3 \\
 b_0 & b_1 & b_2 & b_3
\end{array}\right|=2(\Delta_{01}\Delta_{23}-\Delta_{02}\Delta_{13}+\Delta_{03}\Delta_{12})$$
where 
$$\Delta_{ij}=\left|\begin{array}{cc}
a_i & a_j \\
b_i & b_j
\end{array}\right|$$
Is there any way of expressing the following determinant in a similar way? That is, is there any way of expanding the following determinant as a sum of products of minors of order $2\times 2$ of the form $\Delta_{ij}$?
$$0=\left|\begin{array}{cccc}
 a_0 & a_1 & \cdots & a_n \\
 b_0 & b_1 & \cdots & b_n \\
 \vdots & \vdots & \cdots & \vdots \\
 a_0 & a_1 & \cdots & a_n \\
 b_0 & b_1 & \cdots & b_n 
\end{array}\right|$$
(of course, one must take into account the parity of $n$). I need it in order to find the set of zeroes that defines a projective algebraic variety.
 A: I'm sorry, I now realise I changed the notation so that my indices run $1$ to $n$ and not $0$ to $n$. 
What you have written down is essentially the Laplace expansion of your matrix using $2\times 2$ minors from the top two rows against the complementary $(n-2)\times (n-2)$ minors. With $n=4$ and your symmetry it collapses the way you say. 
If $n=2k$ we can expand by the minors of the first two rows against their complements, and then repeat the process. I think we get the rather horrid
$$
\sum\epsilon(i_1,j_1,\dots,i_k,j_k)\ \Delta_{i_1,j_1}\cdot\Delta_{i_2,j_2}\cdot\dots\cdot\Delta_{i_k,j_k}
$$
where the sum is over all
$$\begin{align}
i_1&<j_1\\
i_2&<j_2 \ \text{and}\ i_2,j_2\in\{1,\dots,2k\}\setminus\{i_1,j_1\}\\
i_3&<j_3 \ \text{and}\ i_3,j_3\in\{1,\dots,2k\}\setminus\{i_1,j_1, i_2,j_2\}\\
\text{and so on}
\end{align}
$$
and $\epsilon(i_1,j_1,\dots,i_k,j_k)$ is the sign of the permutation. 
In fact the distinct terms will each turn up $k!$ times but to write this down I'd need much better notation. 
